Simplify the following expression and state the conditions under which the simplification is valid. You can assume that $z \neq 0$. $q = \dfrac{z^2 - 9z}{z^2 + 8z + 7} \times \dfrac{z + 1}{z - 9} $
Explanation: First factor the quadratic. $q = \dfrac{z^2 - 9z}{(z + 1)(z + 7)} \times \dfrac{z + 1}{z - 9} $ Then factor out any other terms. $q = \dfrac{z(z - 9)}{(z + 1)(z + 7)} \times \dfrac{z + 1}{z - 9} $ Then multiply the two numerators and multiply the two denominators. $q = \dfrac{ z(z - 9) \times (z + 1) } { (z + 1)(z + 7) \times (z - 9) } $ $q = \dfrac{ z(z - 9)(z + 1)}{ (z + 1)(z + 7)(z - 9)} $ Notice that $(z - 9)$ and $(z + 1)$ appear in both the numerator and denominator so we can cancel them. $q = \dfrac{ z(z - 9)\cancel{(z + 1)}}{ \cancel{(z + 1)}(z + 7)(z - 9)} $ We are dividing by $z + 1$ , so $z + 1 \neq 0$ Therefore, $z \neq -1$ $q = \dfrac{ z\cancel{(z - 9)}\cancel{(z + 1)}}{ \cancel{(z + 1)}(z + 7)\cancel{(z - 9)}} $ We are dividing by $z - 9$ , so $z - 9 \neq 0$ Therefore, $z \neq 9$ $q = \dfrac{z}{z + 7} ; \space z \neq -1 ; \space z \neq 9 $